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Question

A particle in SHM follows \( x = 2 \sin (4t + \frac{\pi}{2}) \) (in m). What is its acceleration at \(
t = 0 \, \text{s} \)?

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Explanation

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. Acceleration: a = -ω² x . ω = 4 s⁻¹, x(0) = 2 sin ((π/2)) = 2 × 1 = 2 m . a = -4² × 2 = -16 × 2 = -32 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result -32 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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